Insert Interval -- leetcode

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:

Given intervals [1,3],[6,9], insert and merge [2,5] in as
[1,5],[6,9].

Example 2:

Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as
[1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        auto iter = intervals.begin();
        while (iter != intervals.end() &&
            iter->end < newInterval.start)
            ++iter;

        if (iter != intervals.end()) {
            auto iend = iter;
            while (iend != intervals.end() &&
                 newInterval.end >= iend->start)
                ++iend;

            if (iter != iend) {
                newInterval.start = min(newInterval.start, iter->start);
                --iend;
                newInterval.end = max(newInterval.end, iend->end);
                ++iend;

                iter = intervals.erase(iter, iend);
            }
        }

        intervals.insert(iter, newInterval);
        return intervals;
    }
};

该算法在leetcode上实际执行时间为16ms。

基本思路为，

在数组中找到和待插入的区间有重叠的所以区间 [iter, iend)， 注，左闭右开。

若存在，合并区间信息到待插入区间。并删除原有的重叠区间。

若不存在，则直接插入。

所要注意的事项为，用范围删除，即先找到范围，再整体删除。

如果找到一个区间，就删除一个。 在运行时间上将不能被AC。